I need to apply change of variables to my problem. I do know explicitly the followings:
$$ x = \phi(\xi,\eta) $$ $$ y = \psi(\xi,\eta) $$
Obtaining the reverse of this is very cumbersome. To calculate the first derivatives:
$$ \frac{\partial (\cdot)}{\partial x} = \frac{\partial \xi}{\partial x} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial \eta}{\partial x} \frac{\partial (\cdot)}{\partial \eta} $$
$$ \frac{\partial (\cdot)}{\partial y} = \frac{\partial \xi}{\partial y} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial \eta}{\partial y} \frac{\partial (\cdot)}{\partial \eta} $$
Since I explicitly know $\frac{\partial x}{\partial \xi}$, $\frac{\partial x}{\partial \eta}$,$\frac{\partial y}{\partial \xi}$,$\frac{\partial y}{\partial \eta}$, I can compute the reverse derivatives by inverting the Jacobian matrix:
$$ [J] = \left[ \matrix{\frac{\partial x}{\partial \xi}&\frac{\partial x}{\partial \eta} \\ \frac{\partial y}{\partial \xi} & \frac{\partial y}{\partial \eta}} \right]$$
The inverse is just simply:
$$ [J]^{-1} = \left[ \matrix{\frac{\partial \xi}{\partial x}&\frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial x} & \frac{\partial \eta}{\partial y}} \right]$$
I need to perform the same operation for second order derivatives, to correctly obtain the derivatives in new variables. So, I end up with Hessian Matrix as shown:
$$ [H] = \left[ \matrix{\frac{\partial^2 x}{\partial \xi^2}&\frac{\partial^2 x}{\partial \xi \partial \eta} \\ \frac{\partial^2 y}{\partial \xi \partial \eta} & \frac{\partial^2 y}{\partial \eta^2}} \right] $$
Here comes the question: if I invert this Hessian Matrix, will it mean the following:
$$ [H]^{-1} = \left[ \matrix{\frac{\partial^2 \xi}{\partial x^2}&\frac{\partial^2 \xi}{\partial x \partial y} \\ \frac{\partial^2 \eta}{\partial x \partial y} & \frac{\partial^2 \eta}{\partial y^2}} \right] $$
Any help would be appreciated, thank you.
After thinking for a while, I think I obtained the necessary answer. It does involve both Jacobian and Hessian matrices.
I know the following:
$$ x = \phi(\xi,\eta) $$ $$ y = \psi(\xi,\eta) $$
Obtaining the reverse of this is very cumbersome. The first and second derivatives are found as follows:
$$ \frac{\partial (\cdot)}{\partial x} = \frac{\partial \xi}{\partial x} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial \eta}{\partial x} \frac{\partial (\cdot)}{\partial \eta} $$
$$ \frac{\partial (\cdot)}{\partial y} = \frac{\partial \xi}{\partial y} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial \eta}{\partial y} \frac{\partial (\cdot)}{\partial \eta} $$
$$ \frac{\partial^2 (\cdot)}{\partial x^2} = \frac{\partial^2 \xi}{\partial x^2} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial^2 \eta}{\partial x^2} \frac{\partial (\cdot)}{\partial \eta} + \left(\frac{\partial \xi}{\partial x}\right)^2 \frac{\partial^2 (\cdot)}{\partial \xi^2} + \left(\frac{\partial \eta}{\partial x}\right)^2 \frac{\partial^2 (\cdot)}{\partial \eta^2} + 2 \left(\frac{\partial \xi}{\partial x}\right)\left(\frac{\partial \eta}{\partial x}\right) \frac{\partial^2 (\cdot)}{\partial \xi \partial \eta} $$
$$ \frac{\partial^2 (\cdot)}{\partial y^2} = \frac{\partial^2 \xi}{\partial y^2} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial^2 \eta}{\partial y^2} \frac{\partial (\cdot)}{\partial \eta} + \left(\frac{\partial \xi}{\partial y}\right)^2 \frac{\partial^2 (\cdot)}{\partial \xi^2} + \left(\frac{\partial \eta}{\partial y}\right)^2 \frac{\partial^2 (\cdot)}{\partial \eta^2} + 2 \left(\frac{\partial \xi}{\partial y}\right)\left(\frac{\partial \eta}{\partial y}\right) \frac{\partial^2 (\cdot)}{\partial \xi \partial \eta} $$
$$ \frac{\partial^2 (\cdot)}{\partial x \partial y} = \frac{\partial^2 \xi}{\partial x \partial y} \frac{\partial (\cdot)}{\partial \xi} + \frac{\partial^2 \eta}{\partial x \partial y} \frac{\partial (\cdot)}{\partial \eta} + \left(\frac{\partial \xi}{\partial x}\frac{\partial \xi}{\partial y}\right) \frac{\partial^2 (\cdot)}{\partial \xi^2} + \left(\frac{\partial \eta}{\partial x}\frac{\partial \eta}{\partial y}\right) \frac{\partial^2 (\cdot)}{\partial \eta^2} + \left(\frac{\partial \xi}{\partial x} \frac{\partial \eta}{\partial y} + \frac{\partial \xi}{\partial y} \frac{\partial \eta}{\partial x} \right) \frac{\partial^2 (\cdot)}{\partial \xi \partial \eta} $$
To save space, I will denote derivatives as subscripts. If we construct matrix to relate derivatives in $x,y$ to $\xi,\eta$:
$$ \left[ \matrix{\partial_x \\ \partial_y \\ \partial_{xx} \\ \partial_{yy} \\ \partial_{xy} }\right] = \left[ \matrix{ \xi_x && \eta_x && 0 && 0 && 0 \\ \xi_y && \eta_y && 0 && 0 && 0 \\ \xi_{xx} && \eta_{xx} && \xi_x^2 && \eta_x^2 && 2 \xi_x \eta_x \\ \xi_{yy} && \eta_{yy} && \xi_y^2 && \eta_y^2 && 2 \xi_y \eta_y \\ \xi_{xy} && \eta_{xy} && \xi_x \xi_y && \eta_x \eta_y && \xi_x \eta_y + \xi_y \eta_x}\right] \left[ \matrix{\partial_\xi \\ \partial_\eta \\ \partial_{\xi\xi} \\ \partial_{\eta\eta} \\ \partial_{\xi\eta} }\right] $$
Here, by making analogy with Jacobian matrix to relate first-order derivatives to each other, the derived matrix relates both first-order and second-order derivatives to each other. Since I explicitly know the followings:
$$ x_\xi, x_\eta, x_{\xi \xi}, x_{\eta \eta}, x_{\xi \eta}, y_\xi, y_\eta, y_{\xi \xi}, y_{\eta \eta}, y_{\xi \eta} $$
It means that I know the inverse relation. Calculating the inverse matrix, then taking the inverse of it, would yield the values I am seeking for which is the matrix above.